Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $q = \dfrac{3z^2 + 27z}{z^2 + 4z - 45} \times \dfrac{-5z^2 + 15z + 50}{-3z^2 + 18z} $
Answer: First factor out any common factors. $q = \dfrac{3z(z + 9)}{z^2 + 4z - 45} \times \dfrac{-5(z^2 - 3z - 10)}{-3z(z - 6)} $ Then factor the quadratic expressions. $q = \dfrac {3z(z + 9)} {(z - 5)(z + 9)} \times \dfrac {-5(z - 5)(z + 2)} {-3z(z - 6)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {3z(z + 9) \times -5(z - 5)(z + 2) } { (z - 5)(z + 9) \times -3z(z - 6)} $ $q = \dfrac {-15z(z - 5)(z + 2)(z + 9)} {-3z(z - 5)(z + 9)(z - 6)} $ Notice that $(z - 5)$ and $(z + 9)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {-15z\cancel{(z - 5)}(z + 2)(z + 9)} {-3z\cancel{(z - 5)}(z + 9)(z - 6)} $ We are dividing by $z - 5$ , so $z - 5 \neq 0$ Therefore, $z \neq 5$ $q = \dfrac {-15z\cancel{(z - 5)}(z + 2)\cancel{(z + 9)}} {-3z\cancel{(z - 5)}\cancel{(z + 9)}(z - 6)} $ We are dividing by $z + 9$ , so $z + 9 \neq 0$ Therefore, $z \neq -9$ $q = \dfrac {-15z(z + 2)} {-3z(z - 6)} $ $ q = \dfrac{5(z + 2)}{z - 6}; z \neq 5; z \neq -9 $